Proof that $\lim_{x \to \infty} \frac{x^{a}}{b^{x}}=0 $ I have to prove that $$\lim_{x \to \infty} \frac{x^{a}}{b^{x}}
$$ 
equals zero for all $a$ of $\mathbb{N}$ and $|b|>1$
The suggestion is to start by finding $$\lim_{x \to \infty} \frac{x^{1}}{b^{x}}
$$ 
which I have done. But what I get is $\frac{x}{b^{x}}$ and since the limit of $b^{x}$ is defined as $0$ I don't know how to continue...
 A: Since $\lim_{x\to\infty}x^a=\infty$ and $\lim_{x\to\infty}b^x=\infty$, you can to use L'Hospital's Rule:
\begin{align}
\lim_{x\to\infty}\frac{x^a}{b^x} = \lim_{x\to\infty}\frac{d(x^a)/dx}{d(b^x)/dx} = \lim_{x\to\infty}\frac{ax^{a-1}}{\ln(b) b^x}
\end{align}
Notice, if $a=1$ then the limit in the numerator is just $1$ while the denominator still goes to infinity, so that the limit of the ratio goes to zero. To complete the proof just notice that for any $a \in \mathbb{N}$ you can just repeat the above step $a$ times until you obtain the same result.
Edit: The above proof work for $b>1$. For the case in which $b<-1$ notice that 
$$ \frac{d(b^x)}{dx} = (\ln(-b)+\pi i)b^x$$
the limit of which is $-\infty$. So an analogous argument works for that case.
A: If you have done 
$$\lim_{x\to \infty} \frac{x}{b^x} = 0, $$
for all $b >1$, then 
$$\lim_{x\to \infty} \frac{x^n}{b^x} = \lim_{x\to \infty} \left(\frac{x}{(\sqrt[n]b)^x}\right)^n = \left( \lim_{x\to \infty}\frac{x}{(\sqrt[n]b)^x}\right)^n =0$$
as $\sqrt[n] b$ is still $>1$. Not sure what's going on when $b<-1$. 
